In Exercises 61–66, find the solution set for each system by grap...

Question

In Exercises 61–66, find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.

Accepted Answer

Hello, today we're going to be solving the system of equations by graphing both of the given equations. So let's go ahead and label the first equation as equation one. And let's go ahead and label the second equation as equation two. Looking at the first equation, the first equation is given to us as X squared over four plus Y squared over nine is equal to one. Now, this is an equation of an ellipse. And one thing to note is that the value underneath the Y value is greater than the value underneath the X value. That means that the equation of the ellipse has the major axis of the Y axis. And that means that the rough shape of this ellipse is going to have the major vertices along the Y axis and the minor vertices along the X axis. So we just need to find the values of the major and minor axis. Well, we're gonna represent the minor axis with the variable B and we're gonna represent the major axis as the variable A. Now, in order to get the A values, we're going to set A squared equal to the larger value in the denominator which in this case is nine. And we're going to set B squared equal to the smaller denominator, which is four. Solving for A squared, we can take the square root of both sides. And that is going to give us A is equal to plus or minus three. And taking the square root of both sides of the B equation is going to give us B is equal to plus or minus two. That means the rough graph of the ellipse is going to have the major vertices at zero, comma three and zero, comma negative three and is going to have the minor vertices at two, comma zero and negative two comma zero. So this is going to be the rough shape of the first equation. Now we just need to figure out the graph of the second equation while the second equation is given to us as three X minus two, Y is equal to six. This isn't the standard form of the equation of the line. And so what we need to do is we need to solve for the Y variable. In order to do that, we can subtract three X to both sides of the equation. And that is going to give us negative two, Y is equal to negative three X plus six. Then to solve for Y, we're gonna divide both sides of this equation by negative two. And that is going to give us Y is equal to 3/2 X minus three. So what this means is that this is the equation of a line where the Y intercept is negative three and it has a slope of 3/2. So if we try to plot this on the ellipse that we drew earlier, the Y intercept of this line is going to be at negative three which also intercepts with one of the major vertices of the ellipse. Then the slope is positive 3/2. So the slope is gonna travel up three units and to the right two units. And that is going to intercept with one of the minor vertices which is going to be at two comma zero. And the rough shape of the graph of the line is gonna look something like this. Now, since we know that the equations of the two graphs intercept at two given points zero, comma negative three and two comma zero, that means that the solutions to this system of equations is going to be zero, comma negative three and negative two comma zero. I'm sorry, positive two comma zero. But now we need to verify that these are the correct solutions to the systems of equations. In order to do that, we need to plug in both of the points to each of the equations and see whether they produce a positive value or not. So recall that the first equation is X squared over four plus Y squared over nine is equal to one. We're gonna start by plugging in the first point which is zero, comma negative three. This means that X is equal to zero and Y is equal to negative three. Since X is equal to 00, divided by anything is going to give us zero. So we have zero plus a negative three squared which is positive 9/9 is equal to one 9/ is going to simplify to give us one and one plus zero is just one. So plugging in the 0.0 comma negative three to the first equation is going to give us one is equal to one which is true. Then we want to check the other point for the first equation which is two comma zero. If we plug in two comma zero into the first equation, we get 4/4 plus zero is equal to +14 divided by four is going to simplify to give us one and one plus zero is equal to one. So we are left with one is equal to one, which is a true statement. So this verifies that the set of solutions satisfies the first equation. Now we want to check to see if it satisfies the second equation. Now the second equation was given to us as three X minus two Y is equal to six. If we plug in the 1st 0.0 comma negative three, what we are left with is zero minus two times negative three which is positive six is equal to 60 plus six is going to give us six. And what we are left with is six is equal to six, which is a true statement. Then plugging in the point to comma zero, that is going to give us six minus zero is equal to six and six minus zero is going to give us six, leaving us with six is equal to six. And that is a true statement. So this verifies that the set of solutions satisfies both of the given equations. And we know the shape of the graph of the equations which was roughly drawn here with that being said the solution to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

In Exercises 61–66, find the solution set for each system by graphing both of the system's equations in the same rectangular coordinate system and finding points of intersection. Check all solutions in both equations.

Key Concepts

Systems of Equations

Graphing Linear Equations

Checking Solutions

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